Assumptions - LINE

• Linearity. The relationship between y and x is linear.
• Independence. The errors \epsilon are independent.
• Normal. The errors \epsilon are normally distributed.
• Equal Variance of errors \epsilon. At each value of x, the variance of y is the same i.e. homoskedasticity, or constant variance.

How do we check the assumptions?

Recall that the linear model is a deterministic straight line equation y = c + mx plus some random noise \epsilon:

Y_i = \beta_0 + \beta_1 x + \epsilon

• If the only source of variation in y is \epsilon, then we can check our assumptions by just looking at the residuals \hat{\epsilon}.

# simulate example data
set.seed(340)
x <- runif(8, 0, 30)
y <- 5 * x + rnorm(8, 0, 40)
df <- data.frame(x, y)

# fit linear model, add residual vertical lines as arrows
mod <- lm(y ~ x, data = df)
p1 <- ggplot(df, aes(x, y)) +
  geom_point() +
  geom_segment(aes(xend = x, yend = fitted(mod)),
    arrow = arrow(length = unit(0.2, "cm")),
    color = "royalblue"
  ) +
  labs(x = "x", y = "y")

p1 +
  geom_smooth(method = "lm", se = FALSE, color = "firebrick") +
  annotate("text",
    x = 6.3, y = -6, size = 7,
    label = expression(hat(epsilon[i])), colour = "royalblue"
  ) +
  annotate("text",
    x = 5.6, y = 25, size = 7,
    label = expression(hat(y[i])), colour = "forestgreen"
  ) +
  annotate("text",
    x = 5.6, y = -36, size = 7,
    label = expression(y[i]), colour = "firebrick"
  ) +
  theme_classic()

Another way to look at residuals

For a given ‘x’, there is a range of ‘y’ values that are possible.

• e.g. For a given parent height, there is a range of child heights that are possible.

1-step

• Residuals vs. Fitted: check for linearity, equal variance.
• Q-Q Residuals: check for normality.
• Scale-Location: check for equal variance (standardised).
• Residuals vs. Leverage: check for outliers (influential points).

par(mfrow = c(2, 2)) # need to do this to get 4 plots on one page
plot(fit)

Assumption: Linearity

• Residuals vs. fitted plot looks at the relationship between the residuals and the fitted values.
• If the relationship is linear:
  • Residuals should be randomly scattered around the horizontal axis.
  • The red line should be reasonably straight.
• Could also look at a scatterplot of x and y!

plot(fit, which = 1)

Examples

Linear Models with R (Faraway 2005, p59)

Assumption: Normality

• Q-Q plot looks at the distribution of the residuals against a normal distribution function (the dotted line).
• Sometimes, a histogram is still useful to see the shape of the distribution.

par(mfrow = c(1, 2))
plot(fit, which = 2)
hist(rstandard(fit))

Assumption: Normality

• If normally distributed, the points should follow the red line.
• Deviation from the red line is common in the tails (i.e. the ends), but not in the middle.

Examples

set.seed(915)
x <- rnorm(100)
y <- 2 + 5 * x + rchisq(100, df = 2)
df <- data.frame(x, y)
fit_eg <- lm(y ~ x, data = df)
par(mfrow = c(1, 2))
plot(fit_eg, which = 2)
hist(rstandard(fit_eg))

Equal variances

• Look at the scale-location plot.
• If variances are equal, the points should be randomly scattered around the horizontal axis.
• The red line should be more or less horizontal.

plot(fit, which = 3)

Equal variances

• If variances are not equal we may see:
  • A funnel shape, where the points are more spread out at the ends than in the middle. Sometimes also called “fanning”.
  • Patterns in the scale-location plot, such as a curve or a wave, indicating that the variance is changing.
• Look at the red line for a general trend, but don’t depend on it too much.

set.seed(915)
x <- rnorm(100)
y <- 2 + 5 * x^2 + rchisq(100, df = 2)
df <- data.frame(x, y)
fit_eg <- lm(y ~ x, data = df)
plot(fit_eg, which = 3)

Outliers

• Leverage is a measure of how far away the predictor variable is from the mean of the predictor variable.
• The Residuals vs Leverage plot shows the relationship between the residuals and the leverage of each point.
• Cook’s distance is a measure of how much the model would change if a point was removed.

In general, points with high leverage and high Cook’s distance are considered outliers.

plot(fit, which = 5)

Example of an influential outlier

set.seed(1028)
x <- rnorm(30)
y <- 2 + 2 * x

y[25] <- y[25] + 40 # Add an outlier

fit_eg <- lm(y ~ x)

par(mfrow=c(1,2))
plot(fit_eg, which = 5)

plot(x, y)
abline(fit_eg, col = "red")

fit_eg2 <- lm(y[-25] ~ x[-25])
abline(fit_eg2, col = "blue", lty = 3)

Points that exceed the dashed line (which appears once they approach the Cook’s distance), are likely to influence the model. These outliers should be removed.

e.g. this is of a perfect line with one extreme outlier – the line of best fit deviates because of a single point!

What can we do if the assumptions aren’t met?

It depends…

…which assumption is not met and the type of data i.e. circumstances.

• If data is non-linear, try a transformation of the response variable y, from light to extreme:

  • root: \sqrt{y} or \sqrt{y+1} if y contains zeros
  • log: \log(y) or \log(y+1) if y contains zeros
  • inverse: \frac{1}{y} or \frac{1}{y+1} if y contains zeros

• If residuals are not normally distributed, try a transformation of the response variable y first, otherwise transform the predictor variable x. Both can be done at the same time.

• If equal variances assumption is not met, same as above.

• If outliers are present, try removing them, or transforming the response variable y.

What if transformation doesn’t work?

If the assumptions are still not met after trying the above, you can try:

• Using a different type of regression e.g. logistic regression, non-linear regression
• Using a different model e.g. machine learning.
• Using a non-parametric test.

Back to Galton - model assumptions are met

Now what?

Interpreting the output

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

• Call: the model formula
• Residuals: distribution of the residuals

Interpreting the output

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

Interpreting the output

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

• Pr: the p-value
  • (Intercept)/Pr: the p-value of the y-intercept is not meaningful.
  • parent/Pr: is parent a significant predictor to the model?

Interpreting the output

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

• We can also use the Estimate values to write the equation of the regression line: \widehat{child} = 23.94153 + 0.64629 \cdot parent

• For every one-inch increase in the parent height, the child height is predicted to increase by 0.64629 inches.

• e.g. if a parent is 70 inches, how tall will the child be? 23.94153 + 0.64629 \cdot 70 = 68.5 inches (174 cm).

Interpreting the output

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

• Residual standard error: the standard deviation of the residuals.
  • Interpretation: the average amount that the response will deviate from the true regression line.
• degrees of freedom: the number of observations minus the number of parameters being estimated. Used in hypothesis testing and calculating the standard error of the regression coefficients.

Interpreting the output

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

• Multiple R-squared: the proportion of variance (0-1) explained by the model (for simple linear regression).
• Adjusted R-squared: the proportion of variance (0-1) explained by the model, adjusted for the number of predictors (for multiple linear regression).
• Ranges from 0 to 1; R² = 1 is a perfect fit.
• “The proportion of variance in the response that is explained by parent: 21.05%.”

Interpreting the output

Call:
lm(formula = child ~ parent, data = Galton)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.8050 -1.3661  0.0487  1.6339  5.9264 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 23.94153    2.81088   8.517   <2e-16 ***
parent       0.64629    0.04114  15.711   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 2.239 on 926 degrees of freedom
Multiple R-squared:  0.2105,    Adjusted R-squared:  0.2096 
F-statistic: 246.8 on 1 and 926 DF,  p-value: < 2.2e-16

• F-statistic: the ratio of the variance explained by predictors, and the residual variance (variance not explained by predictors).
  • Also known as the partial F-test between the full model and the intercept-only (null) model.
• p-value: the probability that the F-statistic is greater than the observed value under the null hypothesis.
  • In a simple linear regression, the p-value for the slope coefficient is the same as the p-value for the F-statistic.

Hypothesis testing

How does our null (H_0: \beta_1=0) model compare to the linear (H_0: \beta_1 \neq 0) model?

In simple linear regression, the p-value for the slope coefficient is the same as the p-value for the F-statistic.

null_model <- Galton %>%
  lm(child ~ 1, data = .) %>%
  augment(Galton)
lin_model <- Galton %>%
  lm(child ~ parent, data = .) %>%
  augment(Galton)
models <- bind_rows(null_model, lin_model) %>%
  mutate(model = rep(c("Null model", "SLR model"), each = nrow(Galton)))

ggplot(data = models, aes(x = parent, y = child)) +
  geom_smooth(
    data = filter(models, model == "Null model"),
    method = "lm", se = FALSE, formula = y ~ 1, size = 0.5
  ) +
  geom_smooth(
    data = filter(models, model == "SLR model"),
    method = "lm", se = FALSE, formula = y ~ x, size = 0.5
  ) +
  geom_segment(
    aes(xend = parent, yend = .fitted),
    arrow = arrow(length = unit(0.1, "cm")),
    size = 0.3, color = "darkgray"
  ) +
  geom_point(alpha = .2) +
  facet_wrap(~model) +
  xlab("Parent height (in)") +
  ylab("Child height (in)")

What are we testing?

• The null model is a model with no predictors, i.e. y = \beta_0 + \epsilon
• The alternative model is a linear model with one predictor, i.e. y = \beta_0 + \beta_1 x + \epsilon
• We use the t-test to compare the two models:

t = \frac{estimate - 0}{Standard\ error} = \frac{\hat{\beta}_1 - 0}{SE(\hat{\beta}_1)} where SE(\hat{\beta}_1) is the standard error of the slope estimate:

SE(\hat{\beta}_1) = \frac{\hat{\sigma}}{\sqrt{\sum_{i=1}^n (x_i - \bar{x})^2}}

Reporting

Refresher on earlier steps

Galton collected data on the average height of both parents and their children. There wasa moderate, positive, linear relationship (r = 0.46) between parent and child height.

Can we rely on our model results? Assumptions and hypothesis?

We fitted a linear model to predict child height from parent height, and model assumptions were met. The model was statistically significant (R² = 0.21, F(1, 926) = 246.84, p < .001), hence the null hypothesis was rejected in favour of the linear model. The effect of parent height is statistically significant and positive (\beta = 0.65, t(926) = 15.71, p < .001).

Equation? Inference?

For every one-inch increase in parent height, child height is predicted to increase by 0.65 inches ($ = 23.94 + 0.65 parent$). The average height of both parents explains 21.05% of the variance in child height – there is an effect but there are clearly more factors at play.

Fun fact

• Galton’s key finding was that children of tall parents are not as tall, children of short parents are not as short.

\widehat{child} = 23.94153 + 0.64629 \cdot parent

• e.g. if average parent height is 60 inches (152 cm), how tall will the child be? 23.94153 + 0.64629 \cdot 60 = 62.7 inches (159 cm).

• e.g. if average parent height is 75 inches (191 cm), how tall will the child be? 23.94153 + 0.64629 \cdot 75 = 72.4 inches (184 cm).

• The height of children appeared to regress towards the population mean, i.e. the concept of regression to the mean

• Hence the Galton is credited with coining the term regression (and also correlation, percentile, median, etc.)

Explore

Read the data:

library(readxl) # load the readxl package

alligator <- read_excel(path = "data/ENVX1002_Lecture_wk10_data.xlsx", 
  sheet = "Alligator") # read in the data

What does the data look like?

str(alligator)

tibble [25 × 2] (S3: tbl_df/tbl/data.frame)
 $ Length: num [1:25] 58 61 63 68 69 72 72 74 74 76 ...
 $ Weight: num [1:25] 28 44 33 39 36 38 61 54 51 42 ...

Plot

plot(x = alligator$Length, y = alligator$Weight, 
  xlab = "Length (cm)", ylab = "Weight (kg)")

library(ggplot2) # load the ggplot2 package
ggplot(data = alligator, aes(x = Length, y = Weight)) +
  geom_point() +
  labs(x = "Length (cm)", y = "Weight (kg)")

Plot residual diagnostics

To check assumptions, we need to fit the model first, then plot the model.

fit <- lm(formula = Weight ~ Length, data = alligator)
par(mfrow = c(2, 2)) # set up a 2 x 2 grid for plots
plot(fit)

Check assumptions

Is the relationship linear?

plot(fit, which = 1)

If the linearity assumption is not met, there is no reason to validate the model since it is no longer suitable for the data.

Dealing with non-linearity: transform the data

library(patchwork)

p1 <- ggplot(data = alligator, aes(x = Length, y = Weight)) +
  geom_point(size = 3) +
  labs(x = "Length (cm)", y = "Weight (kg)", title = "Original") +
  geom_smooth(se = FALSE, linetype = 2)

p2 <- ggplot(data = alligator, aes(x = Length, y = sqrt(Weight))) +
  geom_point(size = 3) +
  labs(x = "Length (cm)", y = "sqrt[Weight (kg)]", title = "Square root") +
  geom_smooth(se = FALSE, linetype = 2)

p3 <- ggplot(data = alligator, aes(x = Length, y = log(Weight))) +
  geom_point(size = 3) +
  labs(x = "Length (cm)", y = "log[Weight (kg)]", title = "Natural log") +
  geom_smooth(se = FALSE, linetype = 2)

p4 <- ggplot(data = alligator, aes(x = Length, y = log10(Weight))) +
  geom_point(size = 3) +
  labs(x = "Length (cm)", y = "log10[Weight (kg)]", title = "Log base 10") +
  geom_smooth(se = FALSE, linetype = 2)

p1 + p2 + p3 + p4

Natural log transformation – Check assumptions again

fit <- lm(formula = log(Weight) ~ Length, data = alligator)
par(mfrow = c(2, 2)) # set up a 2 x 2 grid for plots
plot(fit)

Interpretation

summary(fit)

Call:
lm(formula = log(Weight) ~ Length, data = alligator)

Residuals:
      Min        1Q    Median        3Q       Max 
-0.289266 -0.079989  0.000933  0.102216  0.288491 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 1.335335   0.131394   10.16 5.63e-10 ***
Length      0.035416   0.001506   23.52  < 2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 0.1493 on 23 degrees of freedom
Multiple R-squared:  0.9601,    Adjusted R-squared:  0.9583 
F-statistic:   553 on 1 and 23 DF,  p-value: < 2.2e-16

• Length is a statistically significant predictor of log(Weight) (p < .001).
• The model explains a statistically significant and large proportion (96%) of variance (R² = 0.96, F(1, 23) = 553, p < .001)
• For every 1 cm increase in Length, log(Weight) increases by 0.0354, Weight increases by e^{0.0354} and Weight increases by approximately 3.54%

Percent change with ln transformation

Interpreting as a percent change can be more meaningful - it can be done with any log transformation (substitute e below for 10 or any other base), but the quick approximation only works with natural log transformations.

If y has been transformed with a natural log (log(y)), for a one-unit increase in x the percent change in y (not log(y)) is calculated with:

\Delta y \% = 100 \cdot (e^{\beta_1}-1)

If \beta_1 is small (i.e. -0.25 < \beta_1 < 0.25), then: e^{\beta_1} \approx 1 + \beta_1. So \Delta y \% \approx 100 \cdot \beta_1.